Signal Processing and Linear Systems-B.P.Lathi copy

Note t hat yot b eing t he z ero input component jt

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Unformatted text preview: own below: A2 + 4A + 40 = (A2 + 4A + 4) + 36 = (A + 2)2 + (6)2 = (A + 2 - j6)(A + 2 + j 6) llO 2 T ime-Domain A nalysis o f C ontinuous-Time S ystems where c and (J are arbitrary constants to be determined from the initial conditions yo(O) = 2 a nd yo(O) = 16.78. Differentiation of Eq. (2.12a) yields yo(t) = _ 2ce- 2t cos (6t + (J) 6ce- 2t sin (6t + Ii) (2.12b) S etting t = 0 in Eqs. (2.12a) and (2.12b), and then substituting initial conditions, we o btain 2 =ccosli - 16.78 = - 2ccos Ii - 6csin Ii Solution of these two simultaneous equations in two unknowns ccos ccos (J = 2 (J and csin csin (J = - 3.463 Squaring and t hen adding the two sides of the above equations yields 2 = (2)2 + ( -3.464)2 = 16 ==> c = 4 c Next, dividing (2.13b) by (2.13a); t hat is dividing csin t an (J = - 3 463 2 and (J Therefore by ccos (J yields (2.13a) (2.13b) yields = t an- 1 (-3 2 463) = -1 yo(t) = 4 e- 2t cos (6t Figure B .llc shows the plot of yo(t). o (J (J C omputer E xample C 2.1 Find the r oots of polynomial >-2 - 1) • + 4>- + 40 a =[l 4 40]; r = r oots(a) o r --2.0000 + 6 .0000i - 2.0000 - 6.0000i 0 C omputer E xample C 2.2 For an LTIC system specified by the differential equation (D2 ( b) ( e) y O=dsolve('D2y+4*Dy+3*y=0','y(0)=3','Dy(0)=-7','t') yO = 2 *exp(-3*t)+exp(-t) y O=dsolve('D2y+4*Dy+4*y=O' , 'y(O)=3', ' Dy(O)=-7', ' t') yO = 3 *exp(-2*t)-exp(-2t)*t y O=dsolve('D2y+4*Dy+40*y=O' , 'y(O)=3' , 'Dy(0)=-7', ' t') yO = 3 *exp(-2*t)*eos(6*t)-1/6*exp(-2t)*sin(6*t) 0 6. E xercise E 2.1 F ind t he z ero-input r esponse o f a n LTIC s ystem described by (D+ 5)y(t) = f (t) i f t he i nitial condition is y(O) = 5. Answer: yo(t) = 5 e- 5t t:::: 0 'V 6. E xercise E 2.2 Solve (D2 + 2D) yo(t) = 0 i f Yo(O) = 1 a nd !io(O) = 4. Hint: T he c haracteristic roots a re 0 a nd - 2. Answer: yo(t) = 3 - 2e- 2t t :::: 0 'V S ystem R esponse t o I nternal C onditions: Z ero-Input R esponse 111 Practical Initial Conditions and the meaning of 0 - and 0 + I n E xample 2.1 t he i nitial c onditions YolO) a nd yolO) were supplied. I n p racticl\l p roblems, we m ust d erive such conditions from t he p hysical s ituation. F or instance, i n a n R LC c ircuit, we m ay b e g iven t he c onditions, s uch as i nitial c apacitor v oltages, a nd i nitial i nductor c urrents, e tc. F rom t his i nformation, we need t o d erive yolO), yolO), . .. for t he d esired variable a s d emonstrated i n t he n ext e xample. I n m uch o f o ur d iscussion, t he i nput is a ssumed t o s tart a t t = 0, u nless o therwise mentioned. Hence, t = 0 is t he r eference p oint. T he c onditions i mmediately b efore t = 0 ( just b efore t he i nput is applied) a re t he c onditions a t t = 0 -, a nd t hose i mmediately a fter t = 0 ( just a fter t he i nput is applied) a re t he c onditions a t t = 0 + ( compare t his w ith t he h istorical t ime f rame B .C. a nd A .D.). I n p ractice, we a re likely t o k now t he i nitial c onditions a t t = 0 - rather t han a t t = 0+. T he t wo s ets o f conditions a re g enerally different, a lthough i n s ome cases t hey m ay b e i dentical. W e a re d ealing w ith t he t otal r esponse y (t), w hich c onsists o f two c omponents; t he z ero-input...
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